Ram can complete the work in 10 hours, Ajit in 12 hours and Suresh in 15 hours. All of them began the work together, but Ram had

The area of a rectangular living room is calculated using length multiplied by width

This area is stated to be 40 square meters (area = 40 sq. meters)

Let the width be x meters

Length will be X+3 meters

Area = Length x Width

40 = X * (X+3)

40 = X^2 + 3X

We solve this quadratic equation

Thus, x = 5 or x= -8

As width cannot be negative, we discard x =-8

Hence, x = 5 is the solution

Width of the room = 5 meters

Answer:

Option C: 0.28

Detailed explanation:

This situation presents a binomial probability distribution.

We need to determine the likelihood that at least 2 thumbtacks land point up out of 5 tossed ones. This can be expressed as;

P(X ≥ 2) = P(2) + P(3) + P(4) + P(5)

Referring to the histogram;

P(5) = 0.02

P(4) = 0.02

P(3) = 0.05

P(2) = 0.19

Consequently;

P(X ≥ 2) = 0.19 + 0.05 + 0.02 + 0.02

P(X ≥ 2) = 0.28

Response:

a. 55 cars

b. 25 cars

Detailed explanation:

Let’s denote the quantity of cars with stereo systems as N(ss), those with air conditioning as N(ac), and those with sunroofs as N(sr).

We find that:

N(ss) = 30

N(ac) = 30

N(sr) = 40

N(ss and ac and sr) = 15

N(at least two) = 30

a.

To calculate how many cars possess at least one feature (N(at least one) or N(ss or ac or sr)), we apply:

N(ss or ac or sr) = N(ss) + N(ac) + N(sr) - N(ss and ac) - N(ss and sr) - N(ac and sr) + N(ss and ac and sr)

N(ss or ac or sr) = 30 + 30 + 40 - (N(at least two) + 2*N(ss and ac and sr)) + 15

Substituting, we find N(ss or ac or sr) = 30 + 30 + 40 - (30 + 2*15) + 15 = 55

b.

For those cars that have exactly one feature, we have:

N(only one) = N(at least one) - N(at least two)

N(only one) = 55 - 30 = 25

∠ROT=160°

∠SOT=100°

Now we calculate ∠SOR as follows: ∠SOR = ∠ROT - ∠SOT

∠SOR = 160° - 100°

∠SOR = 60°

It is stated that the angles ROQ, QOS, and POT all have the same measure.

Thus, ∠SOQ + ∠QOR = 60°

Since ∠SOQ equals ∠QOR, we can express this as:

2∠SOQ = 60°

From which we find ∠SOQ = 60° ÷ 2

∠SOQ = 30°

Also, ∠POT = ∠SOQ = ∠ROQ = 30°

Given that,

Julia completes a 20-mile bike ride in 1.2 hours.

The distance Julia covers is 20 miles and her time taken is 1.2 hours.

Therefore, Julia's speed = = 16.67 mph

Katie finishes the same 20-mile ride in 1.6 hours.

Katie’s distance is 20 miles and her time is 1.6 hours.

Hence, Katie's speed = = 12.5 mph

To determine how much faster Julia rides compared to Katie, subtract Katie’s speed from Julia’s speed.

Thus, 16.67 mph minus 12.5 mph equals 4.17 mph, approximately 4.2 mph.

Consequently, Julia cycles 4.2 mph faster than Katie.